# derivative very tricky : need guide

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• Jul 30th 2012, 11:45 PM
rcs
derivative very tricky : need guide
How are these pattered in the rule of derivatives?
Find the derivative of dy/dx of the relations:

1.) x^3y+ 2y^4 - x^4 = 2xy

and

2.) x^2 = x+2y/ x - 2y

thanks a lot
• Jul 31st 2012, 12:45 AM
jgv115
Re: derivative very tricky : need guide
If I want to simplify $\frac{d(y)}{dx}$, I obviously can't do it that easily as the variables are wrong.

However, I can apply the chain rule: $\frac{d(y)}{dx}.\frac{dx}{dy}=\frac{d(y)}{dy}$

Solving for $\frac{d(y)}{dx}$, I get $\frac{d(y)}{dx}=\frac{d(y)}{dy}.\frac{dy}{dx}$

Evaluate $\frac{d(y)}{dy}$ which is $1$. Therefore $\frac{d(y)}{dx}= \frac{dy}{dx}$

Now try apply these principles in your question
• Jul 31st 2012, 01:35 AM
rcs
Re: derivative very tricky : need guide
ill try sir... ill be back and let you see my answer also... hope i can get it right

thanks
• Jul 31st 2012, 01:40 AM
emakarov
Re: derivative very tricky : need guide
Quote:

Originally Posted by jgv115
If I want to simplify $\frac{d(y)}{dx}$, I obviously can't do it that easily as the variables are wrong.

However, I can apply the chain rule: $\frac{d(y)}{dx}.\frac{dx}{dy}=\frac{d(y)}{dy}$

Solving for $\frac{d(y)}{dx}$, I get $\frac{d(y)}{dx}=\frac{d(y)}{dy}.\frac{dy}{dx}$

Evaluate $\frac{d(y)}{dy}$ which is $1$. Therefore $\frac{d(y)}{dx}= \frac{dy}{dx}$

Now try apply these principles in your question

To be honest, I don't understand this. What is $\frac{d(y)}{dx}$ and how is it different from $\frac{dy}{dx}$?

OP, it seems that you need to differentiate an implicit function. Obviously, you need to read how to do this, for example, in Wikipedia. Once you learn the method and read several finished examples, please describe the exact difficulty you are having with this particular problem. The thing is that this forum is not a tutorial service and you can't expect people to explain completely new material to you. The forum is here to help those who know the material but struggle to apply it in a concrete situation.
• Jul 31st 2012, 03:36 AM
rcs
Re: derivative very tricky : need guide
thank you
• Jul 31st 2012, 03:37 AM
Deveno
Re: derivative very tricky : need guide
i'll give you an example, perhaps you'll see how it works:

if x2+y2 = 1, find dy/dx.

step 1) differentiate both sides with respect to x:

2x(dx/dx) + 2y(dy/dx) = 0 but dx/dx = 1, so:

2x + 2y(dy/dx) = 0

step 2) solve for dy/dx:

2y(dy/dx) = -2x

dy/dx = -2x/2y = -x/y
• Jul 31st 2012, 04:40 PM
x3bnm
Re: derivative very tricky : need guide
So this is called implicit differentiation. You can differentiate both sides with respect to $x$ where $y$ is a function of $x$ and $\frac{dx}{dx} = 1$:

\begin{aligned} x^3 y + 2 y^4 -x^4 & = 2xy \\ x^3 \frac{dy}{dx} + 3x^2 y + 8y^3 \frac{dy}{dx} - 4x^3 & = 2y + 2x \frac{dy}{dx} \\ \\ [\text{ Now you gather dy/dx one side and isolating dy/dx }] \\ \\ \frac{dy}{dx} & = \frac{2y - 3x^2 y + 4 x^3}{x^3 + 8 y^3 - 2x} \end{aligned}

For the second one:

\begin{aligned} x^2 & = x + \frac{2y}{x} - 2y \\ 2x & = 1 + \frac{x \frac{2 dy}{dx} - 2y}{x^2} - 2 \frac{dy}{dx} \\ \\ [\text{Same as before gather dy/dx together and simplifying using Algebra you get }] \\ \\ \frac{dy}{dx} & = \frac{2x^3 -x^2 + 2y}{2x - 2x^2} \end{array}

You can check it to confirm here for the first equation:

differentiation x^3 y +2 y^4 -x^4 - 2xy = 0 - Wolfram|Alpha

And for the second one:

differentiation x^2 - x - 2y/x +2y = 0 - Wolfram|Alpha

Hope this helps.
• Jul 31st 2012, 05:28 PM
rcs
Re: derivative very tricky : need guide
a million of thanks for the solution sir... the best Helper of this site... --- rcs ----
• Jul 31st 2012, 06:11 PM
x3bnm
Re: derivative very tricky : need guide
Quote:

Originally Posted by rcs
a million of thanks for the solution sir... the best Helper of this site... --- rcs ----

• Sep 2nd 2012, 06:32 PM
rcs
Re: derivative very tricky : need guide
Quote:

Originally Posted by x3bnm
So this is called implicit differentiation. You can differentiate both sides with respect to $x$ where $y$ is a function of $x$ and $\frac{dx}{dx} = 1$:

\begin{aligned} x^3 y + 2 y^4 -x^4 & = 2xy \\ x^3 \frac{dy}{dx} + 3x^2 y + 8y^3 \frac{dy}{dx} - 4x^3 & = 2y + 2x \frac{dy}{dx} \\ \\ [\text{ Now you gather dy/dx one side and isolating dy/dx }] \\ \\ \frac{dy}{dx} & = \frac{2y - 3x^2 y + 4 x^3}{x^3 + 8 y^3 - 2x} \end{aligned}

\begin{aligned} x^2 & = x + \frac{2y}{x} - 2y \\ 2x & = 1 + \frac{x \frac{2 dy}{dx} - 2y}{x^2} - 2 \frac{dy}{dx} \\ \\ [\text{Same as before gather dy/dx together and simplifying using Algebra you get }] \\ \\ \frac{dy}{dx} & = \frac{2x^3 -x^2 + 2y}{2x - 2x^2} \end{array}

You can check it to confirm here for the first equation:

differentiation x^3 y +2 y^4 -x^4 - 2xy = 0 - Wolfram|Alpha

And for the second one:

differentiation x^2 - x - 2y/x +2y = 0 - Wolfram|Alpha

Hope this helps.

i found it confusing .. how did you get this
For the second one:

x^2 = 1 + 2y/x - 2y sir how did you this ?

because it is supposed to be in number 2: the equation is x^2 = (x+2y)/(x-2y)

thanks
• Sep 2nd 2012, 08:05 PM
Prove It
Re: derivative very tricky : need guide
x3bnm read the question AS YOU WROTE IT. If you had used the brackets in the correct place in the first place, you would have been given a correct solution.

Anyway, you now know how to differentiate implicitly, so I suggest you try it out for yourself with the correct question.
• Sep 3rd 2012, 05:05 AM
rcs
Re: derivative very tricky : need guide
Quote:

Originally Posted by Prove It
x3bnm read the question AS YOU WROTE IT. If you had used the brackets in the correct place in the first place, you would have been given a correct solution.

Anyway, you now know how to differentiate implicitly, so I suggest you try it out for yourself with the correct question.

is it possible to get the derivative like ,
x^2(x-2y) = x + 2y sir?
multiplying both side by x - 2y of the equation, before taking the derivative
• Sep 4th 2012, 03:00 PM
x3bnm
Re: derivative very tricky : need guide
Quote:

Originally Posted by rcs
is it possible to get the derivative like ,
x^2(x-2y) = x + 2y sir?
multiplying both side by x - 2y of the equation, before taking the derivative

Yes the new solution is given below:

\begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\ 2x =& \frac{(x - 2y)(1 + 2 \frac{dy}{dx}) - (x+2y)(1 - 2 \frac{dy}{dx})}{(x-2y)^2} \\ 2x =& \frac{x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx}}{(x -2y)^2} \\ 2x(x - 2y)^2 =& x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx} \\ 2x(x - 2y)^2 + 4y =& 4x \frac{dy}{dx} \\ \frac{dy}{dx} =& \frac{2x(x - 2y)^2 + 4y}{4x} \\ \frac{dy}{dx} =& \frac{x(x-2y)^2 + 2y}{2x} \\ \frac{dy}{dx} =& \frac{x^3 - 4x^2 y + 4 xy^2 + 2y}{2x} \text{.................[after simplification] } \end{align*}

Also you can check the answer at:

differentiation x^2 = (x+2y)/(x-2y) - Wolfram|Alpha
• Sep 7th 2012, 05:50 AM
rcs
Re: derivative very tricky : need guide
Quote:

Originally Posted by x3bnm
Yes the new solution is given below:

\begin{align*} x^2 =& \frac{(x+2y)}{(x-2y)} \\ 2x =& \frac{(x - 2y)(1 + 2 \frac{dy}{dx}) - (x+2y)(1 - 2 \frac{dy}{dx})}{(x-2y)^2} \\ 2x =& \frac{x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx}}{(x -2y)^2} \\ 2x(x - 2y)^2 =& x + 2x\frac{dy}{dx} - 2y -4y \frac{dy}{dx} -x + 2x\frac{dy}{dx} -2y +4y\frac{dy}{dx} \\ 2x(x - 2y)^2 + 4y =& 4x \frac{dy}{dx} \\ \frac{dy}{dx} =& \frac{2x(x - 2y)^2 + 4y}{4x} \\ \frac{dy}{dx} =& \frac{x(x-2y)^2 + 2y}{2x} \\ \frac{dy}{dx} =& \frac{x^3 - 4x^2 y + 4 xy^2 + 2y}{2x} \text{.................[after simplification] } \end{align*}